Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x-2y &= 8 \\ x+y &= 2\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $x = {-y+2}$ Substitute this expression for $x$ in the first equation. $8({-y + 2}) - 2y = 8$ $-8y + 16 - 2y = 8$ Simplify by combining terms, then solve for $y$ $-10y + 16 = 8$ $-10y = -8$ $y = \dfrac{4}{5}$ Substitute $\dfrac{4}{5}$ for $y$ in the top equation. $8x-2( \dfrac{4}{5}) = 8$ $8x-\dfrac{8}{5} = 8$ $8x = \dfrac{48}{5}$ $x = \dfrac{6}{5}$ The solution is $\enspace x = \dfrac{6}{5}, \enspace y = \dfrac{4}{5}$.